1
$\begingroup$

Let $I$ be an open interval and let $c\in I$. Let $f:I\rightarrow\mathbb{R}$ be continuous and define $g:I\rightarrow\mathbb{R}$ by $g(x)=\left|f(x)\right|$.

Prove that if $g$ is differentiable at $c$, then $f$ is also differentiable at $c$.

The hint given was to use Caratheodory's Lemma. Since $g$ is differentiable at $c$, there exists a function $\phi:I\rightarrow\mathbb{R}$ that is continuous at $c$ such that

$$g(x)-g(c)=\left|f(x)\right|-\left|f(c)\right|=\phi(x)\cdot(x-c)$$

I need some help on how to proceed.

$\endgroup$
3
$\begingroup$

Hint: Try by separating the three cases: $f(c)>0$ , $f(c)<0$ and $f(c)=0$.

  1. If $f(c)>0$, by the continuity of $f$ at $c$ there is a neighborhood $J\subset I$ of $c$ such that $f_{/J}>0$.
  2. Use same argument if $f(c)<0$.
  3. If $f(c)=0$, show that $\phi(x)\to 0$ (as $x\to c$), i.e $g'(c)=0$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.